TheNightKing wrote:
Jayesh24 wrote:
Hello,
Can I solve it as (5C2 + 6C1 * 5C1)/11C2?
Jayesh24Can you explain reasoning behind each term and how you arrived at this?
May be then someone can help you!
Hello,
Jayesh24. The short answer is
yes. You will arrive at the same answer by solving the problem in the proposed manner. That is,
From 5 green hats, you could select 2 in successive selections: 5C2 (which comes to 10) or
From two successive selections, you could select 1 red hat and 1 green hat: 6C1 * 5C1 (in either order, which will come to 6 * 5, or 30)
Since there are two possibilities to select
at least 1 green hat, you create an
or condition, meaning you add the independent probabilities: (5C2) + (6C1 * 5C1) (or 10 + 30)
Finally, out of 11 total hats, you are selecting 2: 11C2 (which comes to 55) and
since probability is given by a desired outcome divided by all possible outcomes,
\(\frac{(5C2 + 6C1 * 5C1)}{11C2}\)
is correct. Or, if you prefer,
\(\frac{(10 + 6 * 5)}{(55)}\), which becomes \(\frac{40}{55}\), or \(\frac{8}{11}\) when reduced.
This is a roundabout way of getting the answer, but a valid method nonetheless. Well done.
- Andrew
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